package com.sxkiler.demo.hard;

import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import java.util.*;
import com.sxkiler.demo.model.*;

/**
basic-calculator-iv=基本计算器 IV
<p>给定一个表达式&nbsp;<code>expression</code>&nbsp;如&nbsp;<code>expression = "e + 8 - a + 5"</code>&nbsp;和一个求值映射，如&nbsp;<code>{"e": 1}</code>（给定的形式为&nbsp;<code>evalvars = ["e"]</code> 和&nbsp;<code>evalints = [1]</code>），返回表示简化表达式的标记列表，例如 <code>["-1*a","14"]</code></p>

<ul>
	<li>表达式交替使用块和符号，每个块和符号之间有一个空格。</li>
	<li>块要么是括号中的表达式，要么是变量，要么是非负整数。</li>
	<li>块是括号中的表达式，变量或非负整数。</li>
	<li>变量是一个由小写字母组成的字符串（不包括数字）。请注意，变量可以是多个字母，并注意变量从不具有像&nbsp;<code>"2x"</code>&nbsp;或&nbsp;<code>"-x"</code>&nbsp;这样的前导系数或一元运算符&nbsp;。</li>
</ul>

<p>表达式按通常顺序进行求值：先是括号，然后求乘法，再计算加法和减法。例如，<code>expression = "1 + 2 * 3"</code>&nbsp;的答案是 <code>["7"]</code>。</p>

<p>输出格式如下：</p>

<ul>
	<li>对于系数非零的每个自变量项，我们按字典排序的顺序将自变量写在一个项中。例如，我们永远不会写像 <code>&ldquo;b*a*c&rdquo;</code> 这样的项，只写 <code>&ldquo;a*b*c&rdquo;</code>。</li>
	<li>项的次数等于被乘的自变量的数目，并计算重复项。(例如，<code>"a*a*b*c"</code> 的次数为 4。)。我们先写出答案的最大次数项，用字典顺序打破关系，此时忽略词的前导系数。</li>
	<li>项的前导系数直接放在左边，用星号将它与变量分隔开(如果存在的话)。前导系数 1 仍然要打印出来。</li>
	<li>格式良好的一个示例答案是&nbsp;<code>["-2*a*a*a", "3*a*a*b", "3*b*b", "4*a", "5*c", "-6"]</code>&nbsp;。</li>
	<li>系数为 0 的项（包括常数项）不包括在内。例如，&ldquo;0&rdquo; 的表达式输出为 []。</li>
</ul>

<p>&nbsp;</p>

<p><strong>示例：</strong></p>

<pre><strong>输入：</strong>expression = "e + 8 - a + 5", evalvars = ["e"], evalints = [1]
<strong>输出：</strong>["-1*a","14"]

<strong>输入：</strong>expression = "e - 8 + temperature - pressure",
evalvars = ["e", "temperature"], evalints = [1, 12]
<strong>输出：</strong>["-1*pressure","5"]

<strong>输入：</strong>expression = "(e + 8) * (e - 8)", evalvars = [], evalints = []
<strong>输出：</strong>["1*e*e","-64"]

<strong>输入：</strong>expression = "7 - 7", evalvars = [], evalints = []
<strong>输出：</strong>[]

<strong>输入：</strong>expression = "a * b * c + b * a * c * 4", evalvars = [], evalints = []
<strong>输出：</strong>["5*a*b*c"]

<strong>输入：</strong>expression = "((a - b) * (b - c) + (c - a)) * ((a - b) + (b - c) * (c - a))",
evalvars = [], evalints = []
<strong>输出：</strong>["-1*a*a*b*b","2*a*a*b*c","-1*a*a*c*c","1*a*b*b*b","-1*a*b*b*c","-1*a*b*c*c","1*a*c*c*c","-1*b*b*b*c","2*b*b*c*c","-1*b*c*c*c","2*a*a*b","-2*a*a*c","-2*a*b*b","2*a*c*c","1*b*b*b","-1*b*b*c","1*b*c*c","-1*c*c*c","-1*a*a","1*a*b","1*a*c","-1*b*c"]
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ol>
	<li><code>expression</code> 的长度在&nbsp;<code>[1, 250]</code>&nbsp;范围内。</li>
	<li><code>evalvars, evalints</code> 在范围&nbsp;<code>[0, 100]</code>&nbsp;内，且长度相同。</li>
</ol>

 */
public class basicCalculatorIV {
    

    class Solution {
        public List<String> basicCalculatorIV(String param0,String[] param1,Integer[] param2) {
            return null;
        }
    }

    @Test
    public void test(){
        Solution solution = new Solution();
        /**
        "e + 8 - a + 5"
["e"]
[1]
        */
        //int [] num1 = new int[]{1,3};
        //int [] num2 = new int[]{2};
        //Assertions.assertEquals(solution.{{questionName}}(num1,num2),2);
    }
}

